Newbie select problem

[chancho]

Hello all, I'm new to this list. Hope that everyone had a nice Christmas Eve.

My problem is with the code below:

--------- begin code --------------

$result = mysql_query( "SELECT * FROM test_me WHERE ( col1 = '$comp_lgn_name' AND col2 = '$comp_lgn_pswrd' )" );
mysql_close( $link );

if ( $result )
{
print "successful";
}
else
header ( "Location: ../comp_lgn_error.html" );
exit;

-------- code end -------------------

The $comp_lgn_name and $comp_lgn_pswrd are two variables that are submitted via form by the user. The query compares the variables to the info in a MySQL database table, but it returns successful everytime regardless of whether the info that user enters matches the info in the database. Any suggestions... I apologize for the ignorant question, but I am a beginner. Thanks for any replies and MERRY X-MAS to all.

[[deleted]]

mysql_query() executes a query and returns a resource-identiefer if teh query worked, or FALSE if the query failed. This sais nothing about how many results the query returned.

You may want to read this:

http://www.hvt-automation.nl/yapf/faq.php?cmd=viewitem&itemid=12

vincent
http://www.hvt-automation.nl/yapf/

[Anon]

OK...

this is basic... check out php.net for the mysql functions, it shouldnt take any more reading than that to get a good understanding of them, no I'm not going to direct u to some archaic manual for this simple question like the last guy... instead, a nice example 🙂

$query = "SELECT * FROM $user_table WHERE username = '$user' AND pass = '$pass'";
$do = mysql_query($query);
if (mysql_num_rows($do) > 0)
{
// found a result
print "correct login...";
} else {
// nope
print "incorrect!";
}

[[deleted]]

lol

By all means, give him a nice example without any explenation of how it works.

If you bothered to actually read the page that I linked to, you'd have found a detailed description of how to retreive data from a database, not some (poorly written) example that is not even correct:

First, if that query returns more than one result for a user/password combination, that would mean that there are two peope with the same username and password, and it should not allow a logon.

Second, you don't check for errors in the query itself.

[chancho]

Thanks to both Vincent and Jaykay for getting me on the right track. I am still having a slight problem though. If I input the correct "login" then the "member_name and pswrd" print statements are executed, despite any incorrect pswrd. Now, if I input an incorrect "member_name" the else statement is never executed. Any suggestions.

$link = mysql_connect( "localhost", $user, $pass );

if ( ! $link )
die ( "Couldn't connect to MySQL" );
else
{
mysql_select_db( $db, $link )
or die ( "Couldn't open database $db: ".mysql_error( $link ) );

$a_query = "SELECT * FROM ismembercmp WHERE member_name = '$cl_name' AND PASSWORD( pswrd = '$cl_pswrd' )";

if ( !$result = mysql_query( $a_query, $link ) )
{
	print "SQL query failed:";
    print mysql_error( $link )."<br>";
    print $a_query."<br>";
}
else if ( mysql_num_rows( $result ) > 0 )
{
	print mysql_result($result,0,"member_name");
	print "<br>";
	print mysql_result($result,0,"pswrd");
}
else
{
	fopen ( "http://localhost/jotr/comp_lgn_error.html", "r" );
}
mysql_close( $link );

}

[chancho]

Disregard my last post, figured it out. But, is this a correct statement to re-direct someone to another page:

fopen ( "http://localhost/jotr/comp_lgn_error.html", "r" );

Again, thanks for the help.

[Anon]

Nyet; you want

header("Location:http://localhost/jets/comp_lgn_error.html");

Or whatever your real URL is 🙂

Note: this will only work if you haven't already sent anything to the browser (even blank lines) - if you do you'll get an error saying "headers already sent".

And I offer this link:

http://www.php.net/docs.php

in the spirit of "Give a man a fish and he'll eat for a day. Give a man a stick of dynamite and he'll be eating fish for a week."